$\overline{AC} = 9$ $\overline{AB} = {?}$ $A$ $C$ $B$ $?$ $9$ $ \sin( \angle ABC ) = \frac{3\sqrt{10} }{10}, \cos( \angle ABC ) = \frac{ \sqrt{10}}{10}, \tan( \angle ABC ) = 3$
Solution: $\overline{AC}$ is the opposite to $\angle ABC$ $\overline{AB}$ is the hypotenuse (note that it is opposite the right angle) SOH CAH TOA We know the opposite side and need to solve for the hypotenuse so we can use the sin function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}} = \frac{9}{\overline{AB}} $ $ \overline{AB}=\frac{9}{\sin( \angle ABC )} = \frac{9}{\frac{3\sqrt{10} }{10}} = 3\sqrt{10}$